# Sum matrix elements group by indices in Python

I have two matrix (same row and column): one with float values, which are grouped by indices in the other matrix. As a result, I want a dictionary or a list with the sums of the elements for each index. Indices always start at 0.

``````A = np.array([[0.52,0.25,-0.45,0.13],[-0.14,-0.41,0.31,-0.41]])
B = np.array([[1,3,1,2],[3,0,2,2]])

RESULT = {0: -0.41, 1: 0.07, 2: 0.03, 3: 0.11}
``````

I found this solution, but I'm searching for a faster one. I'm working with matrix with 784 x 300 cells and this algorithm takes ~28ms to complete.

``````import numpy as np

def matrix_sum_by_indices(indices,matrix):
a = np.hstack(indices)
b = np.hstack(matrix)
sidx = a.argsort()
split_idx = np.flatnonzero(np.diff(a[sidx])>0)+1
out = np.split(b[sidx], split_idx)
return [sum(x) for x in out]
``````

If you can help me find a better and plain solution to this problem, I'll be grateful!

EDIT: I made a mistake, time to complete is ~8ms in a 300*10 matrix, but ~28ms in a 784x300.

EDIT2: My `A` elements are float64, so `bincount` give me ValueError.

The numpy_indexed package has efficient and simple solutions to this problem (disclaimer: I am its author):

``````import numpy_indexed as npi
keys, values = npi.group_by(B.flatten()).sum(A.flatten())
``````

You can make use of `bincount` here:

``````a = np.array([[0.52,0.25,-0.45,0.13],[-0.14,-0.41,0.31,-0.41]])
b = np.array([[1,3,1,2],[3,0,2,2]])

N = b.max() + 1
id = b + (N*np.arange(b.shape[0]))[:, None] # since you can't apply bincount to a 2D array
np.sum(np.bincount(id.ravel(), a.ravel()).reshape(a.shape[0], -1), axis=0)
``````

Output:

``````array([-0.41,  0.07,  0.03,  0.11])
``````

As a function:

``````def using_bincount(indices, matrx):
N = indices.max() + 1
id = indices + (N*np.arange(indices.shape[0]))[:, None] # since you can't apply bincount to a 2D array
return np.sum(np.bincount(id.ravel(), matrx.ravel()).reshape(matrx.shape[0], -1), axis=0)
``````

Timings on this sample:

``````In [5]: %timeit using_bincount(b, a)
31.1 µs ± 1.74 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)

In [6]: %timeit matrix_sum_by_indices(b, a)
61.3 µs ± 2.62 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)

In [88]: %timeit scipy.ndimage.sum(a, b, index=[0,1,2,3])
54 µs ± 218 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
``````

(`scipy.ndimage.sum` should be faster on much larger samples)

The following solution, relying on scipy.ndimage.sum is highly optimized for speed:

``````import numpy as np
A = np.array([[0.52,0.25,-0.45,0.13], [-0.14,-0.41,0.31,-0.41]])
B = np.array([[1,3,1,2], [3,0,2,2]])
import scipy.ndimage
print(scipy.ndimage.sum(A, B, index=[0,1,2,3]))
``````

You may have to work a little for having the `index` parameter be exactly what you want. It is the list of the indices you want to get in the result. Maybe the following will be a good starting point:

``````print(scipy.ndimage.sum(A,B, index=np.unique(B)))
``````

but if you know by advance the list of all indices, it will be more efficient to hard-code it here.